Answer : The partial pressure of
at equilibrium is, 1.0 × 10⁻⁶
Explanation :
The partial pressure of
= ![1.0\times 10^{-2}atm](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-2%7Datm)
The partial pressure of
= ![2.0\times 10^{-4}atm](https://tex.z-dn.net/?f=2.0%5Ctimes%2010%5E%7B-4%7Datm)
The partial pressure of
= ![2.0\times 10^{-4}atm](https://tex.z-dn.net/?f=2.0%5Ctimes%2010%5E%7B-4%7Datm)
![K_p=4.2\times 10^{-9}](https://tex.z-dn.net/?f=K_p%3D4.2%5Ctimes%2010%5E%7B-9%7D)
The balanced equilibrium reaction is,
![2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)](https://tex.z-dn.net/?f=2HBr%28g%29%5Crightleftharpoons%20H_2%28g%29%2BBr_2%28g%29)
Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴
At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)
The expression of equilibrium constant
for the reaction will be:
![K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%28p_%7BH_2%7D%29%28p_%7BBr_2%7D%29%7D%7B%28p_%7BHBr%7D%29%5E2%7D)
Now put all the values in this expression, we get :
![4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}](https://tex.z-dn.net/?f=4.2%5Ctimes%2010%5E%7B-9%7D%3D%5Cfrac%7B%282.0%5Ctimes%2010%5E%7B-4%7D%2Bp%29%282.0%5Ctimes%2010%5E%7B-4%7D%2Bp%29%7D%7B%281.0%5Ctimes%2010%5E%7B-2%7D-2p%29%5E2%7D)
![p=-1.99\times 10^{-4}](https://tex.z-dn.net/?f=p%3D-1.99%5Ctimes%2010%5E%7B-4%7D)
The partial pressure of
at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶
Therefore, the partial pressure of
at equilibrium is, 1.0 × 10⁻⁶